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It a hard problem, but I know how to solve it (I just learned binomial formula)

Ok the first step is to find combinations 5C3= 5!/(5-3)!*3!=5*4/2=10
Second is to find the probability of all wanted events, 1/2^3 (for all three it will be 1/8)
Third is to find the probability of all unwanted events, 1/2^2 (1/4)

now you have to multiply all these numbers

10/8*4

the answer is 10/32 or 5/16

P.S. sorry for a lame explanation, but it's very hard to explain (check the attached document)

Hi you answer would have been correct if the problem would have been "Three Heads" but I think you overlooked Consecutive heads...so the asnwer is 3/16.

Saurabh Malpani

Sergey_is_cool wrote:

It a hard problem, but I know how to solve it (I just learned binomial formula)

Ok the first step is to find combinations 5C3= 5!/(5-3)!*3!=5*4/2=10 Second is to find the probability of all wanted events, 1/2^3 (for all three it will be 1/8) Third is to find the probability of all unwanted events, 1/2^2 (1/4)

now you have to multiply all these numbers

10/8*4

the answer is 10/32 or 5/16

P.S. sorry for a lame explanation, but it's very hard to explain (check the attached document)

Hi you answer would have been correct if the problem would have been "Three Heads" but I think you overlooked Consecutive heads...so the asnwer is 3/16.

Saurabh Malpani

Sergey_is_cool wrote:

It a hard problem, but I know how to solve it (I just learned binomial formula)

Ok the first step is to find combinations 5C3= 5!/(5-3)!*3!=5*4/2=10 Second is to find the probability of all wanted events, 1/2^3 (for all three it will be 1/8) Third is to find the probability of all unwanted events, 1/2^2 (1/4)

now you have to multiply all these numbers

10/8*4

the answer is 10/32 or 5/16

P.S. sorry for a lame explanation, but it's very hard to explain (check the attached document)

The question stem asks not just about consecutive heads, but ''at least three'' consecutive heads: this means, of course, the sum of the probobility of 3 consecutive heads, the probability of 4 consecutive heads, and the probability of 5 consecutive heads. So--Himalayan-- you have these possibilies: HHHTT, THHHT, TTHHH, HHHHT, THHHH, and HHHHH. This ends up resulting in a probability of 6/32 = 3/16.

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...